[mizuchaud]

Well, my original question was how many shuffles would it take to get two single color stacks.

I think 99's asking the question if I mean " back to their original poisitions": ie. red on the left and black on the right. I'm just interested in the original question as posed.

Just by doing shuffles and not MATH (yikes) I get an interesting number of shuffles for multiples of 4 chips in stacks (4 and 8 chips in the stacks) to attain single color stacks.

SPOILER BELOW..

I get three shuffles to reattain single colors for a 3 chip stack.

Three shuffles to achieve single colors for a 4 chip stack, but in the wrong positions.

Five shuffles for five chip stacks, but wrong position.

Six shuffles for 6 chips, with wrong positions.

7 for 7

but the kicker is that I get FOUR shuffles for an eight chip stack.

In my primitive head I latched onto the multiples of four as a subset of shuffling numbers.
junford, can you state that as a solution?


I'm hurting my head trying to figure out jdunford's phrasing.
what's the ^ mean?